Single particle states in QFT are defined to be irreducible invariant subspaces under the action of the Poincare group. Such states are generally specified by a spin projection and a momentum. Is it possible for a 1 particle state to have other quantum numbers?

My expectation is no, as existence of any other degrees of freedom would suggest that the state is not irreducible.


1 Answer 1


Hint: colour, electric charge, etc. (keyword: internal symmetry).

Recall that in QM we always label the states of the theory by simultaneously diagonalising a complete set of commuting operators, typically the Hamiltonian and a maximal set of commuting symmetries (conserved charges). These commuting operators become labels on states.

One must therefore ask oneself what is the most general set of symmetries allowed in a certain QFT, and here Coleman-Mandula is essential: this theorem guarantees that the most general set is given by the direct product of Poincaré times a compact Lie algebra. These symmetries are dubbed external and internal respectively.

As these two sets don't mix (they commute), you can diagonalise them independently, and so you can perform the standard Wigner classification, so you end up with 1PS labelled by external quantum numbers: mass, three-momentum, spin and helicity. No other external quantum numbers exist (unless we consider higher dimensional theories, where there are more than one spin quantum number). Internal symmetries give you additional quantum numbers such as electric charge, colour, etc., but OP, as most people do, does no consider these quantum numbers to label different states for a certain 1PS, but different 1PS altogether. In other words, for a fixed 1PS (fixed internal quantum numbers), the only labels are external (mass, spin, etc.).

Note that CM is crucial here: without it it would not be possible in general to split the symmetries into two groups (external and internal). For example, in QFT's with Galilean symmetries, a boost and a translation is in a sense equivalent to a phase rotation, so the three-momentum label is not completely independent from the electric charge quantum number. In general you can have very exotic mixing of internal and external symmetries, and therefore the whole classification of states can be arbitrarily different from what one expects based on the standard relativistic classification.

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    $\begingroup$ I think with the definition given by OP he treats e.g. quarks of different color as different particles. So it all boils down to terminology, and in OP's terminology only momenta and spin are quantum numbers of a 1-particle state. $\endgroup$ Apr 20, 2017 at 7:40
  • $\begingroup$ @SolenodonParadoxus in that case, I'd like to mention another keyword: Coleman-Mandula. I have to leave now, but I might expand on it later on if OP thinks it might be useful for them. $\endgroup$ Apr 20, 2017 at 7:57
  • $\begingroup$ @AccidentalFourierTransform Please do elaborate. Solendon's interpretation of my terminology is correct. $\endgroup$
    – Yly
    Apr 20, 2017 at 8:25
  • $\begingroup$ @Yly AccidentalFourierTransform probably means the Coleman-Mandula theorem. It forbids internal and spacetime symmetries to mix in any way but trivial, provided a certain number of assumptions (quite realistic ones) about scattering amplitudes. It is certainly related, but I disagree that it answers the question. IMO a simple classification of unitary Poincare group irreps is enough to conclude that only momenta&spin label 1-particle states. $\endgroup$ Apr 20, 2017 at 23:29
  • $\begingroup$ +1, thanks for the explanation. Your point is: Wiegner's classification singles out momenta and spin eigenvalues as labeling 1 particle states, while CM is essential for the very existence of such a subspace, right? $\endgroup$ Apr 21, 2017 at 23:39

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